Choose the most correct option. Benzaldehyde does NOT show positive test with
1. Schiff reagent
2. Tollens’ reagent
3. Sodium bisulphate solution
4. Fehling solution

Hint: We know that an aldehyde (R-CO-H) is a functional group where carbonyl carbon is at least bonded to one Hydrogen atom. The chemical formula of benzaldehyde is \[{{\rm{C}}_{\rm{6}}}{{\rm{H}}_{\rm{5}}}{\rm{CHO}}\].

Complete Step by Step Solution:
Let’s understand whether the Schiff reagent gives a a positive test with Schiff reagent or not. The use of Schiff reagents is mainly to distinguish between aldehydes and ketones. Ketones do not undergo reaction with Schiff reagent. But, aldehydes undergo a reaction with Schiff reagent. Therefore, Benzaldehyde gives a positive test with Schiff reagent.
Let’s understand the reaction of Benzaldehyde with Tollen’s reagent. Tollens reagent has a chemical formula of \[{\left( {{\rm{Ag}}\left( {{\rm{NH3}}} \right)} \right)^{{\rm{ + 2}}}}\] It is an oxidising agent of mild nature. When it undergoes reaction with Benzaldehyde, the CO group got oxidised and Ag+ is reduced. And an oxidised aldehyde is obtained which is a radical cation. The reaction of this cation with hydroxide gives a tetrahedral intermediate. Therefore, Tollen's reagent gives a positive test with Benzaldehyde.

Let’s understand the reaction of Benzaldehyde with Sodium bisulphate solution. The reaction of benzaldehyde with Sodium bisulphate results in the formation of a white coloured white precipitate. Therefore, Sodium bisulphate gives a positive test with Benzaldehyde.
Let’s understand the reaction of Benzaldehyde with the Fehling solution. Fehling's solution possesses a \[{\rm{C}}{{\rm{u}}^{{\rm{2 + }}}}\] ion which forms complex with two tartrate ions. Benzaldehyde has no alpha hydrogen and thus the formation of an enolate does not take place. And there, benzaldehyde gives a negative test with Fehling's solution.

Hence, the option (4) is right.

Note: It is to be noted that, Benzaldehyde is the aromatic aldehyde whose structure is \[{{\rm{C}}_{\rm{6}}}{{\rm{H}}_{\rm{5}}}{\rm{CHO}}\].Here, carbonyl group is the electron withdrawing group.